Problem: The equation of a circle is given below. $(x-20)^{2}+(y-0.05)^{2} = 81$ What is its center? $($
Standard equation of the circle A circle is the collection of all points at a distance ${r}$ from a center $({h},{k})$. We can use the Pythagorean theorem to write an equation to relate the center and radius. $({h},{k})$ ${r}$ $x-{h}$ $y-{k}$ $(x, y)$ $\begin{aligned} a^2+b^2&=c^2\\\\ (x - {h})^2 + (y - {k})^2 &= {r}^2 \end{aligned}$ Rewriting the given equation We can rewrite the given equation as: $\begin{aligned}(x-20)^{2}+(y-0.05)^{2} &= 81\\\\ (x - {20})^2 + (y - {0.05})^2 &= {81}\end{aligned}$ Finding the center According to the rewritten equation, we can see that the center of the circle is $({20}, {0.05})$. Finding the radius According to the standard equation of the circle, we get that ${r^2}={81}$. Solving for the radius, we get that $r={\sqrt{81}}={9}$. Summary The circle is centered at $(20,0.05)$. The circle has a radius of $9$ units.